Calculus Examples

$lim_{x \to 2} \frac{4 \ln (2 x - 3)}{2 e^{2 x - 4} - 2}$

Step 1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to 2} 4 \ln (2 x - 3)}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Evaluate the limit.

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Step 1.2.1.1

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{4 lim_{x \to 2} \ln (2 x - 3)}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2.1.2

Move the limit inside the logarithm.

$\frac{4 \ln (lim_{x \to 2} 2 x - 3)}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{4 \ln (lim_{x \to 2} 2 x - lim_{x \to 2} 3)}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2.1.4

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{4 \ln (2 lim_{x \to 2} x - lim_{x \to 2} 3)}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2.1.5

Evaluate the limit of $3$ which is constant as $x$ approaches $2$ .

$\frac{4 \ln (2 lim_{x \to 2} x - 1 \cdot 3)}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{4 \ln (2 \cdot 2 - 1 \cdot 3)}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Multiply $2$ by $2$ .

$\frac{4 \ln (4 - 1 \cdot 3)}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2.3.1.2

Multiply $- 1$ by $3$ .

$\frac{4 \ln (4 - 3)}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2.3.2

Subtract $3$ from $4$ .

$\frac{4 \ln (1)}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2.3.3

The natural logarithm of $1$ is $0$ .

$\frac{4 \cdot 0}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.2.3.4

Multiply $4$ by $0$ .

$\frac{0}{lim_{x \to 2} 2 e^{2 x - 4} - 2}$

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Evaluate the limit.

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Step 1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{0}{lim_{x \to 2} 2 e^{2 x - 4} - lim_{x \to 2} 2}$

Step 1.3.1.2

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{2 lim_{x \to 2} e^{2 x - 4} - lim_{x \to 2} 2}$

Step 1.3.1.3

Move the limit into the exponent.

$\frac{0}{2 e^{lim_{x \to 2} 2 x - 4} - lim_{x \to 2} 2}$

Step 1.3.1.4

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{0}{2 e^{lim_{x \to 2} 2 x - lim_{x \to 2} 4} - lim_{x \to 2} 2}$

Step 1.3.1.5

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{2 e^{2 lim_{x \to 2} x - lim_{x \to 2} 4} - lim_{x \to 2} 2}$

Step 1.3.1.6

Evaluate the limit of $4$ which is constant as $x$ approaches $2$ .

$\frac{0}{2 e^{2 lim_{x \to 2} x - 1 \cdot 4} - lim_{x \to 2} 2}$

Step 1.3.1.7

Evaluate the limit of $2$ which is constant as $x$ approaches $2$ .

$\frac{0}{2 e^{2 lim_{x \to 2} x - 1 \cdot 4} - 1 \cdot 2}$

Step 1.3.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{0}{2 e^{2 \cdot 2 - 1 \cdot 4} - 1 \cdot 2}$

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

Simplify each term.

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Step 1.3.3.1.1

Simplify each term.

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Step 1.3.3.1.1.1

Multiply $2$ by $2$ .

$\frac{0}{2 e^{4 - 1 \cdot 4} - 1 \cdot 2}$

Step 1.3.3.1.1.2

Multiply $- 1$ by $4$ .

$\frac{0}{2 e^{4 - 4} - 1 \cdot 2}$

Step 1.3.3.1.2

Subtract $4$ from $4$ .

$\frac{0}{2 e^{0} - 1 \cdot 2}$

Step 1.3.3.1.3

Anything raised to $0$ is $1$ .

$\frac{0}{2 \cdot 1 - 1 \cdot 2}$

Step 1.3.3.1.4

Multiply $2$ by $1$ .

$\frac{0}{2 - 1 \cdot 2}$

Step 1.3.3.1.5

Multiply $- 1$ by $2$ .

$\frac{0}{2 - 2}$

Step 1.3.3.2

Subtract $2$ from $2$ .

$\frac{0}{0}$

Step 1.3.3.3

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.3.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to 2} \frac{4 \ln (2 x - 3)}{2 e^{2 x - 4} - 2} = lim_{x \to 2} \frac{\frac{d}{d x} [4 \ln (2 x - 3)]}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 2} \frac{\frac{d}{d x} [4 \ln (2 x - 3)]}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.2

Since $4$ is constant with respect to $x$ , the derivative of $4 \ln (2 x - 3)$ with respect to $x$ is $4 \frac{d}{d x} [\ln (2 x - 3)]$ .

$lim_{x \to 2} \frac{4 \frac{d}{d x} [\ln (2 x - 3)]}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 2 x - 3$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $2 x - 3$ .

$lim_{x \to 2} \frac{4 (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [2 x - 3])}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.3.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$lim_{x \to 2} \frac{4 (\frac{1}{u} \frac{d}{d x} [2 x - 3])}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.3.3

Replace all occurrences of $u$ with $2 x - 3$ .

$lim_{x \to 2} \frac{4 (\frac{1}{2 x - 3} \frac{d}{d x} [2 x - 3])}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.4

Combine $\frac{1}{2 x - 3}$ and $4$ .

$lim_{x \to 2} \frac{\frac{4}{2 x - 3} \frac{d}{d x} [2 x - 3]}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.5

By the Sum Rule, the derivative of $2 x - 3$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3]$ .

$lim_{x \to 2} \frac{\frac{4}{2 x - 3} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3])}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.6

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to 2} \frac{\frac{4}{2 x - 3} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 3])}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.7

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 2} \frac{\frac{4}{2 x - 3} (2 \cdot 1 + \frac{d}{d x} [- 3])}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.8

Multiply $2$ by $1$ .

$lim_{x \to 2} \frac{\frac{4}{2 x - 3} (2 + \frac{d}{d x} [- 3])}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.9

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$lim_{x \to 2} \frac{\frac{4}{2 x - 3} (2 + 0)}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.10

Add $2$ and $0$ .

$lim_{x \to 2} \frac{\frac{4}{2 x - 3} \cdot 2}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.11

Combine $\frac{4}{2 x - 3}$ and $2$ .

$lim_{x \to 2} \frac{\frac{4 \cdot 2}{2 x - 3}}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.12

Multiply $4$ by $2$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{\frac{d}{d x} [2 e^{2 x - 4} - 2]}$

Step 3.13

By the Sum Rule, the derivative of $2 e^{2 x - 4} - 2$ with respect to $x$ is $\frac{d}{d x} [2 e^{2 x - 4}] + \frac{d}{d x} [- 2]$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{\frac{d}{d x} [2 e^{2 x - 4}] + \frac{d}{d x} [- 2]}$

Step 3.14

Evaluate $\frac{d}{d x} [2 e^{2 x - 4}]$ .

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Step 3.14.1

Since $2$ is constant with respect to $x$ , the derivative of $2 e^{2 x - 4}$ with respect to $x$ is $2 \frac{d}{d x} [e^{2 x - 4}]$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 \frac{d}{d x} [e^{2 x - 4}] + \frac{d}{d x} [- 2]}$

Step 3.14.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 2 x - 4$ .

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Step 3.14.2.1

To apply the Chain Rule, set $u$ as $2 x - 4$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x - 4]) + \frac{d}{d x} [- 2]}$

Step 3.14.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 (e^{u} \frac{d}{d x} [2 x - 4]) + \frac{d}{d x} [- 2]}$

Step 3.14.2.3

Replace all occurrences of $u$ with $2 x - 4$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 (e^{2 x - 4} \frac{d}{d x} [2 x - 4]) + \frac{d}{d x} [- 2]}$

Step 3.14.3

By the Sum Rule, the derivative of $2 x - 4$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 4]$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 (e^{2 x - 4} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 4])) + \frac{d}{d x} [- 2]}$

Step 3.14.4

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 (e^{2 x - 4} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 4])) + \frac{d}{d x} [- 2]}$

Step 3.14.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 (e^{2 x - 4} (2 \cdot 1 + \frac{d}{d x} [- 4])) + \frac{d}{d x} [- 2]}$

Step 3.14.6

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 (e^{2 x - 4} (2 \cdot 1 + 0)) + \frac{d}{d x} [- 2]}$

Step 3.14.7

Multiply $2$ by $1$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 (e^{2 x - 4} (2 + 0)) + \frac{d}{d x} [- 2]}$

Step 3.14.8

Add $2$ and $0$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 (e^{2 x - 4} \cdot 2) + \frac{d}{d x} [- 2]}$

Step 3.14.9

Move $2$ to the left of $e^{2 x - 4}$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{2 (2 e^{2 x - 4}) + \frac{d}{d x} [- 2]}$

Step 3.14.10

Multiply $2$ by $2$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{4 e^{2 x - 4} + \frac{d}{d x} [- 2]}$

Step 3.15

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{4 e^{2 x - 4} + 0}$

Step 3.16

Add $4 e^{2 x - 4}$ and $0$ .

$lim_{x \to 2} \frac{\frac{8}{2 x - 3}}{4 e^{2 x - 4}}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 2} \frac{8}{2 x - 3} \cdot \frac{1}{4 e^{2 x - 4}}$

Step 5

Simplify terms.

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Step 5.1

Multiply $\frac{8}{2 x - 3}$ by $\frac{1}{4 e^{2 x - 4}}$ .

$lim_{x \to 2} \frac{8}{(2 x - 3) (4 e^{2 x - 4})}$

Step 5.2

Cancel the common factor of $8$ and $4$ .

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Step 5.2.1

Factor $4$ out of $8$ .

$lim_{x \to 2} \frac{4 \cdot 2}{(2 x - 3) (4 e^{2 x - 4})}$

Step 5.2.2

Cancel the common factors.

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Step 5.2.2.1

Factor $4$ out of $(2 x - 3) (4 e^{2 x - 4})$ .

$lim_{x \to 2} \frac{4 \cdot 2}{4 ((2 x - 3) (e^{2 x - 4}))}$

Step 5.2.2.2

Cancel the common factor.

$lim_{x \to 2} \frac{4 \cdot 2}{4 ((2 x - 3) (e^{2 x - 4}))}$

Step 5.2.2.3

Rewrite the expression.

$lim_{x \to 2} \frac{2}{(2 x - 3) (e^{2 x - 4})}$

Step 6

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 lim_{x \to 2} \frac{1}{(2 x - 3) e^{2 x - 4}}$

Step 7

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $2$ .

$2 \frac{lim_{x \to 2} 1}{lim_{x \to 2} (2 x - 3) e^{2 x - 4}}$

Step 8

Evaluate the limit of $1$ which is constant as $x$ approaches $2$ .

$2 \frac{1}{lim_{x \to 2} (2 x - 3) e^{2 x - 4}}$

Step 9

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $2$ .

$2 \frac{1}{lim_{x \to 2} 2 x - 3 \cdot lim_{x \to 2} e^{2 x - 4}}$

Step 10

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$2 \frac{1}{(lim_{x \to 2} 2 x - lim_{x \to 2} 3) \cdot lim_{x \to 2} e^{2 x - 4}}$

Step 11

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \frac{1}{(2 lim_{x \to 2} x - lim_{x \to 2} 3) \cdot lim_{x \to 2} e^{2 x - 4}}$

Step 12

Evaluate the limit of $3$ which is constant as $x$ approaches $2$ .

$2 \frac{1}{(2 lim_{x \to 2} x - 1 \cdot 3) \cdot lim_{x \to 2} e^{2 x - 4}}$

Step 13

Move the limit into the exponent.

$2 \frac{1}{(2 lim_{x \to 2} x - 1 \cdot 3) \cdot e^{lim_{x \to 2} 2 x - 4}}$

Step 14

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$2 \frac{1}{(2 lim_{x \to 2} x - 1 \cdot 3) \cdot e^{lim_{x \to 2} 2 x - lim_{x \to 2} 4}}$

Step 15

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \frac{1}{(2 lim_{x \to 2} x - 1 \cdot 3) \cdot e^{2 lim_{x \to 2} x - lim_{x \to 2} 4}}$

Step 16

Evaluate the limit of $4$ which is constant as $x$ approaches $2$ .

$2 \frac{1}{(2 lim_{x \to 2} x - 1 \cdot 3) \cdot e^{2 lim_{x \to 2} x - 1 \cdot 4}}$

Step 17

Evaluate the limits by plugging in $2$ for all occurrences of $x$ .

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Step 17.1

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$2 \frac{1}{(2 \cdot 2 - 1 \cdot 3) \cdot e^{2 lim_{x \to 2} x - 1 \cdot 4}}$

Step 17.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$2 \frac{1}{(2 \cdot 2 - 1 \cdot 3) \cdot e^{2 \cdot 2 - 1 \cdot 4}}$

Step 18

Simplify the answer.

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Step 18.1

Simplify the denominator.

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Step 18.1.1

Multiply $2$ by $2$ .

$2 \frac{1}{(4 - 1 \cdot 3) e^{2 \cdot 2 - 1 \cdot 4}}$

Step 18.1.2

Multiply $- 1$ by $3$ .

$2 \frac{1}{(4 - 3) e^{2 \cdot 2 - 1 \cdot 4}}$

Step 18.1.3

Subtract $3$ from $4$ .

$2 \frac{1}{1 e^{2 \cdot 2 - 1 \cdot 4}}$

Step 18.1.4

Multiply $e^{2 \cdot 2 - 1 \cdot 4}$ by $1$ .

$2 \frac{1}{e^{2 \cdot 2 - 1 \cdot 4}}$

Step 18.1.5

Simplify each term.

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Step 18.1.5.1

Multiply $2$ by $2$ .

$2 \frac{1}{e^{4 - 1 \cdot 4}}$

Step 18.1.5.2

Multiply $- 1$ by $4$ .

$2 \frac{1}{e^{4 - 4}}$

Step 18.1.6

Subtract $4$ from $4$ .

$2 \frac{1}{e^{0}}$

Step 18.1.7

Anything raised to $0$ is $1$ .

$2 (\frac{1}{1})$

Step 18.2

Cancel the common factor of $1$ .

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Step 18.2.1

Cancel the common factor.

$2 (\frac{1}{1})$

Step 18.2.2

Rewrite the expression.

$2 \cdot 1$

Step 18.3

Multiply $2$ by $1$ .

$2$